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  • Lectures on Nuclear and Hadron Physics

    Lectures on Nuclear and Hadron Physics

    J. A. Oller

    Departamento de F́ısica Universidad de Murcia

    July 27, 2011

  • Lectures on Nuclear and Hadron Physics

    Contents

    Outline Lecture III

    1 Introduction

    2 On the σ Resonance

    3 On the ρ Resonance

    4 Algebraic N/D

    5 IAM

    6 Mixing of Scalar Resonances

    7 Large NC 8 Lightest Glueball

  • Lectures on Nuclear and Hadron Physics

    Introduction

    Introduction

    1 The mesonic scalar sector has the vacuum quantum numbers JPC = 0++. Essential for the study of Spontaneous Breaking of Chiral Symmetry

    2 The interactions with these quantum numbers are also essential for the masses of the lightest hadrons, the pseudo-Goldstone bosons (π, K , η), and ratio of light quark masses L. Roca, J.A.O. Eur. Phys. J. A 34 (2007) 371

    3 In this sector meson interactions are really strong Large unitarity loops (infrared enhancements and dynamical generation of resonances) Channels that couple very strongly, e.g. ππ-KK̄ , πη-KK̄ Resonances with large widths and in coupled channels: NO Breit-Wigner form, NO Vector-Meson-Dominance, NO narrow-resonance approximation, etc

    4 Okubo-Zweig-Izuka rule (large NC QCD) has large corrections:

  • Lectures on Nuclear and Hadron Physics

    Introduction

    NO ideal mixing multiplets

    Simple quark model fails

    Large unitarity loops

  • Lectures on Nuclear and Hadron Physics

    σ Resonance

    σ Resonance

    The “lightest” hadronic resonance in QCD√ sσ (MeV) Reference

    469 − i 203 Oller, J.A.O. NPA620(1997)438; (E)NPA652(1999)307

    441+16−8 − i 271+9−12 Caprini,Colangelo,Leutwyler PRL96(2006)132001

    484 ± 14 − i 255 ± 10 Garćıa-Mart́ın, Peláez, Ynduráin PRD76(2007)074034

    456 ± 6 − i 241 ± 7 Albaladejo,J.A.O. PRL101(2008)252002 440+3−3 − i 258+2−3 Z.-H.Guo, J.A.O., PRD in press

  • Lectures on Nuclear and Hadron Physics

    σ Resonance

    Connection of the σ with chiral symmetry

    The lowest order chiral perturbation theory Lagrangian

    L2 = F 2π 4

    Tr ( ∂µU∂

    µU† + U†χ+ χ†U )

    χ = 2B0(s + i p)

    U = exp {

    i √

    2Φ/Fπ

    }

    Φ =

     

    1√ 2 π0 + 1√

    6 η8 π

    + K+

    π− − 1√ 2 π0 + 1√

    6 η8 K

    0

    K− K 0 − 2√

    6 η8

     

  • Lectures on Nuclear and Hadron Physics

    σ Resonance

    The lowest order JPC = 0++ ππ scattering amplitude π(p1)π(p2) → π(p3)π(p4) Mandelstam variables: s = (p1 + p2)

    2 , t = (p1 − p3)2 and u = (p1 − p4)2

    T2 = s − m2π/2

    F 2π

    It has a zero at s = m2π/2 (Adler zero) For higher partial waves ℓ ≥ 1 the Adler zero is trivial. It is located at threshold.

  • Lectures on Nuclear and Hadron Physics

    σ Resonance

    Coming back to the N/D method without LHC:

    T =

    [ ∑

    i

    γi s − si

    + g(s)

    ]−1

    To reproduce the Adler zero ONE CDD is needed

    γ1 = F 2 π

    s1 = m 2 π/2

  • Lectures on Nuclear and Hadron Physics

    σ Resonance

    g(s) is the unitarity loop function

    g(s) = g(s0) − ∫ ∞

    4m2π

    ds ′ ρ(s ′)

    (s ′ − s0)(s ′ − s − iǫ)

    = 1

    (4π)2

    [ a(µ) + log

    m2π µ2

    + σ(s) log σ(s) + 1

    σ(s) − 1

    ]

    Fitting ππ phase shifts

    a(Mρ) = −0.79

    T =

    [ 1

    T2 + g(s)

    ]−1 =

    T2 1 + T2g(s)

    = T2 − T2g(s)T2 + . . .

    On the natural size for a(µ): First Unitarity Correction

    T 22 g(s)

    T2 ∼ a

    16π2 s − m2π/2

    F 2π ∼ a s

    16π2F 2π

    |a| must be O(1). Dynamical Generated Resonance

  • Lectures on Nuclear and Hadron Physics

    σ Resonance

    Pole position of the σ

    • Physical amplitudes have no pole positions in the complex s plane with imaginary part because hermiticity would be violated The eigenvalues of a Hermitian operator like the Hamiltonian must be real. The pole positions for physical amplitudes should be real (bound states) K. Gottfried, Quantum Mechanics, W.A. Benjamin

    • Scattering amplitudes can be extended to unphysical Riemann sheets. There one can find poles corresponding to resonances or antibound states

  • Lectures on Nuclear and Hadron Physics

    σ Resonance

    Multi-valued function:

    p =

    √ s

    4 − m2π

    Physical sheet: ℑ|p(s)| ≥ 0 Second Riemann sheet: ℑ|p(s)| ≤ 0

    +iǫ

    −iǫ

    I-SHEET

    I-SHEET II-SHEET

    II-SHEET

    ℑg(s) = −|p|(s) 8π

    √ s → ℑgII (s) =

    |p|(s) 8π

    √ s

  • Lectures on Nuclear and Hadron Physics

    σ Resonance

    Schwartz’s reflection principle:

    g(s + iǫ) − g(s − iǫ) = 2iℑg(s + iǫ) = −i |p| 4π

    √ s

    g(s + iǫ) − gII (s + iǫ) = −i |p|

    4π √

    s

    gII (s) = g(s) + i |p|

    4π √

    s

    TII (s)= [ T−12 + gII (s)

    ]−1

    TII (s) −1= T (s)−1 + i

    |p| 4π

    √ s

  • Lectures on Nuclear and Hadron Physics

    σ Resonance

    The σ pole position is located in TII (s) Because TII (s

    ∗) = TII (s) ∗ there are indeed two poles at sσ and s∗σ

    Laurent series

    P

    P ′

    sth R

    Physical sheet, s variable

    Because TII (s − iǫ) = T (s + iǫ) the pole P (with negative imaginary part) is effectively much closer to the physical region than the pole P ′ (with positive imaginary part)

  • Lectures on Nuclear and Hadron Physics

    σ Resonance

    TII (s) = γ20

    s − sP︸ ︷︷ ︸ pole

    + γ1 + γ2(s − sP)︸ ︷︷ ︸ non−resonant terms

    + . . .

    The reason why the σ has been traditionally hard to establish experimentally is due to the large contribution of non-resonant terms. Due to the need of cancellation for having the Adler zero.

    sσ = (0.466 − i 0.224)2 GeV2

    γ20 = 5.3 + i 7.7 GeV 2

    γ1 = −8.1 + i 36.9 γ2 = 1.1 + i 0.1 GeV

    2

  • Lectures on Nuclear and Hadron Physics

    σ Resonance

    -50

    0

    50

    100

    150

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    300 400 500 600 700 800 900 1000 1100 1200

    δ11 |T |11 2

    0

    500

    1000

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    300 400 500 600 700 800 900 1000 1100 1200

    00|T | 2

    -100

    -50

    0

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    300

    300 400 500 600 700 800 900 1000 0

    500

    1000

    1500

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    2500

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    3500

    4000

    4500

    300 400 500 600 700 800 900 1000

    background

    ρ

    ρ

    σ

    σ

    BW

    δ00 BW

  • Lectures on Nuclear and Hadron Physics

    σ Resonance

    Breit Wigner formula

    T ∝ e iδ sin δ = sin δ cos δ − i sin δ =

    1

    cot δ − i The partial wave T has a maximum at the resonance position, so that sin δ = ±1 → cot δ = 0

    cot δ = cot δ(ER) + (E − ER)(d cot δ/dE )|E=ER (d cot δ/dE )|E=ER = −2/Γ

    1

    cot δ − i = 1

    −2(E − ER)/Γ − i =

    −Γ/2 E − ER + iΓ/2

  • Lectures on Nuclear and Hadron Physics

    σ Resonance

    If Γ ≪ ER then it is a good approximation

    E + ER ≃ 2ER , s − sR ≃ (E − ER)2ER

    −Γ/2 E − ER + iΓ/2

    ≃ −ΓER s − sR + iΓ ER

    sR = E 2 R

    This is a Breit-Wigner (BW) formula for the amplitude

    It is applied in many phenomenological studies to reproduce data

    BW formula does not work well when typically the resonance is very wide (like the σ) or it is located just on top of a threshold, even more when several channels couple

    BW works very bad for the σ indeed.

  • Lectures on Nuclear and Hadron Physics

    ρ resonance

    The lightest preexisting resonance ρ(770)

    Mρ = 770 MeV, Γρ = 150 MeV

    N/D method without LHC:

    T =

    [ ∑

    i

    γi s − si

    + g(s)

    ]−1

    To reproduce the (Adler) zero at threshold ONE CDD is needed

    γ1 = 6F 2 π

    s1 = 4m 2 π

    a = −6F 2 π

    M2ρ 16π2 ≃ −14.2

  • Lectures on Nuclear and Hadron Physics

    ρ resonance

    |a| ≫ 1 Preexisting Resonance (qq̄ resonance)

    Another argument to appreciate the difference:

    a + log m2π µ2

    = m2π µ̄2

    → µ̄ = µ exp(−a/2)

    with µ ∼ 800 MeV

    µ̄σ = 1.76 GeV

    µ̄ρ = 970 GeV ∼ 1 TeV

    To be compared with the natural scale ΛχPT = 4πFπ ∼ 1.2 GeV

    The ρ pole dominates over the background in the Laurent expansion

    The BW formula for the ρ compares very well with the full result

  • Lectures on Nuclear and Hadron Physics

    ρ resonance

    Pole:

    γ2R s − sR

    Its square root gives the coupling of the resonance to the ππ state |gσ→ππ| = 3.1 GeV |gρ→